解
x²-2x+2=0
x²-2x=-2
2(x³-5/2x²-x+1)-(2x³-x²+1/2)+x²+8x
=(2x³-2x³)+(-5x²+x²+x²)+(-2x+8x)+(2-1/2)
=-3x²+6x+3/2
=-3(x²-2x)+3/2
=-3×(-2)+3/2
=6+3/2
=15/2
已知x²－2x＋2=0,
已知x²－2x＋2=0,
求代數(shù)式2（x³-二分之五x²-x+1）-（2x³-x²+二分之一）+x²+8x的值
求代數(shù)式2（x³-二分之五x²-x+1）-（2x³-x²+二分之一）+x²+8x的值
數(shù)學(xué)人氣：215 ℃時(shí)間：2020-08-16 01:05:16
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